Worked example 6.4: Bullet and block

Question: A bullet of mass $m=12 {\rm g}$ strikes a stationary wooden block of mass $M=5.2 {\rm kg}$ standing on a frictionless surface. The block, with the bullet embedded in it, acquires a velocity of $v=1.7 {\rm m/s}$. What was the velocity of the bullet before it struck the block? What fraction of the bullet's initial kinetic energy is lost (i.e., dissipated) due to the collision with the block?

Answer: Let $u$ be the initial velocity of the bullet. Momentum conservation requires the total horizontal momentum of the system to be the same before and after the bullet strikes the block. The initial momentum of the system is simply $m u$, since the block is initially at rest. The final momentum is $(M+m) v$, since both the block and the bullet end up moving with velocity $v$. Hence,

$$m u = (M+m) v,$$

giving

$$u = \frac{M+m}{m} v = \frac{(0.012+5.2)\times 1.7}{0.012} = 738.4 {\rm m/s}.$$

The initial kinetic energy of the bullet is

$$K_i = \frac{1}{2} m u^2 = 0.5\times 0.012 \times 738.4^2 = 3.2714 {\rm kJ}.$$

The final kinetic energy of the system is

$$K_f = \frac{1}{2} (M+m) v^2 = 0.5\times(0.012+5.2)\times 1.7^2 = 7.53 {\rm J}.$$

Hence, the fraction of the initial kinetic energy which is dissipated is

$$f = \frac{K_i-K_f}{K_i} = \frac{3.2714\times 10^3-7.53}{3.2714\times 10^3} = 0.9977.$$